Question

Ch. 1. Problems 6, 7 & 9

6. The only swimmng pool at a motel is outdoors. It is 5.0 m wide and 12.0 m long. If the weekly
evaporation is 2.35 inches how many gallons of water must be added to the pool if it does not rain? If during the next week the pool still loses two. 35 inches of water to evaporation, even with 20 mm of rainfall, how many liters of water must be added?

Answer 1

946 gal

629 gal

Explanation:

volume that evaporates = LWH

volume = (5.0 m × 100 cm/m × inch / 2.54 cm) × (12.0 m × 100 cm/m × inch / 2.54 cm) × 2.35 inch

volume = 218,550 in.³

volume = 218,550 in.³ × 1 gal / 231 in.³

volume = 946 gallons

You must add 946 gallons.

The loss to evaporation is 2.35 in.

The amount of rain is 20 mm

The net loss is

2.35 in. × 25.4 mm / inch - 20 mm = 39.69 mm

39.69 mm × 1 inch / 25.4 mm = 1.563 in.

The net loss now is 1.563 in. instead of 2.35 in.

We find the fraction of the the previous loss is the loss with rain and multiply by the previous loss.

1.563/2.35 × 946 gal = 629 gal

With rain, then 629 gallons of water must be added.